Problem: $\overline{AC} = 9$ $\overline{BC} = {?}$ $A$ $C$ $B$ $9$ $?$ $ \sin( \angle BAC ) = \frac{ \sqrt{10}}{10}, \cos( \angle BAC ) = \frac{3\sqrt{10} }{10}, \tan( \angle BAC ) = \dfrac{1}{3}$
$\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the adjacent side and need to solve for the opposite side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{\overline{BC}}{9} $ $ \overline{BC}=9 \cdot \tan( \angle BAC ) = 9 \cdot \dfrac{1}{3} = 3$